= Solution
The varying <magnetic flux> through the ring induces an <electromotive force> by <Faraday's law of induction>. Its direction follows <Lenz's law>. With solenoid current $I_s$, ring current $I_r$, ring <self-inductance> $L_r$, <electrical resistance> $R$, and <mutual inductance> $M(z)$,
$$
L_r\dot I_r+RI_r=-M\dot I_s,\qquad F_z=I_sI_r\frac{dM}{dz}.
$$
The ring sits in a nonuniform <magnetic field>: coupling decreases above the solenoid, so $M>0$ and $M'<0$ with a common orientation for the two windings. For $I_s=I_0\cos\omega t$, the in-phase part of the induced ring current is negative:
$$
\langle I_sI_r\rangle=-\frac{I_0^2\omega^2ML_r}{2(R^2+\omega^2L_r^2)},\qquad
\boxed{\langle F_z\rangle=-\frac{I_0^2\omega^2ML_rM'}{2(R^2+\omega^2L_r^2)}>0.}
$$
Thus <alternating-current induction levitation> can balance the ring's weight when $\langle F_z\rangle=mg$. A falling mean lifting <force> with height supplies vertical <stability>. In the strongly inductive limit the two currents are almost opposite: the ring repels the solenoid during both half cycles. Finite <self-inductance> matters; an ideal purely resistive loop would carry a current exactly in quadrature with $I_s$ and give zero average <force> in this model. The source supplies the <Joule heating> in the real ring, and any mechanical work while it rises.
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